Конспект лекций по математическому анализу. Шерстнев А.Н. - 456 стр.

x 2 U . tOGDA W SOOTWETSTWII S 256.1 ODNOZNA^NO OPREDELENO OTOBRAVE-
NIE A00(x) 2 L(E; L(E; F )), NAZYWAEMOE 2-J PROIZWODNOJ OTOBRAVENIQ A.
uDOBNO OTOVDESTWLQTX A00(x) S \LEMENTOM PROSTRANSTWA L(E  E; F ) (SM.
223.13) KAK 2-LINEJNYM OTOBRAVENIEM, DEJSTWU@]IM PO FORMULE
                  A00(x)fh; kg  (A00(x)h)k (h; k 2 E ):
aNALOGI^NO WWODQTSQ PROIZWODNYE BOLEE WYSOKIH PORQDKOW.
   w ZAKL@^ENIE MY PRIWEDEM ANALOG FORMULY tEJLORA, OGRANI^IW[ISX
SLU^AEM OSTATKA W FORME pEANO PRI n = 2, I UKAVEM EE PRIMENENIE K
NAHOVDENI@ DOSTATO^NYH USLOWIJ LOKALXNOGO \KSTREMUMA FUNKCIONALA.
   2. pUSTX W USLOWIQH P. 1 OTOBRAVENIE A00 OPREDELENO I NEPRERYWNO
W U . eSLI fx + th j 0  t  1g  U , TO
      A(x + h) = A(x) + A0(x)h + 12 A00(x)fh; hg + o(khk2) (h ! ):
  tAK KAK A0 DIFFERENCIRUEMO W U , IMEEM
()             A0(x + h) , A0(x) = A00(x)h + o(h) (h ! ):
pRIMENQQ FORMULU nX@TONA-lEJBNICA 259.5 K WEKTOR-FUNKCII
t ! [A(x + th)]0 = A0(x + th)h (0  t  1), IMEEM (S U^ETOM ())
                              Z1
         A(x + h) , A(x) = 0 A0(x + th)h dt
                              Z1
                          =      [A0(x)h + (A00(x)th)h + o(th)h] dt
                               0
                          = A0(x)h + 12 A00(x)fh; hg + r(h);
           Z1
GDE r(h) = 0 o(th) hdt. pOKAVEM, ^TO r(h) = o(khk2) (h ! ). dLQ PROIZ-
WOLXNOGO " > 0 SU]ESTWUET  > 0 TAKOE, ^TO khk <  WLE^ET kok(hhk)k < ",
OTKUDA
             1 
 kZ 1o(th) hdtk  1 
 Z 1ko(th)k khk dt < ":
            khk2 0                 khk2 0
|TO I OZNA^AET, ^TO hlim kr(h)k = 0: >
                      ! khk2

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